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frozen [14]
3 years ago
12

PLEASE SHOW YOUR WORK!!! (solve system of equations by adding, subtracting, or multiplying.)

Mathematics
1 answer:
Aliun [14]3 years ago
5 0
Hey friend, hope I can assist you!
I will solve by elimination <3.
Multiply 3x - 7y = 2 by 2: 6x - 14y = 4
6x - 14y = 4
6x - 9y = 9
5y = 5
Now we have
6x - 14y = 4
5y = 5
Now we want to solve 5y = 5 for y
So simply divide both sides by 5.
5y/5 = 5/5
This gives us one or in other words, y = 1.
Now we want to plug y = 1 into  6x - 14y = 4
So 6x - 14 * 1 = 4
This gives us
6x - 14 = 4
Now add 14 to both sides.
6x - 14 + 14 = 4 + 14
6x = 18
Now divide both sides by 6
6x/6 = 18/6
This gives us 3 so x = 3
Therefore our solutions to this system of equations would be y = 1 and x = 3

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1. The probability of telesales representative making a sale on a customer call is 0.15.
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Answer:

1c

 n = 33

1d

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Step-by-step explanation:

From the question we are told that

   The  probability of telesales representative making a sale on a customer call is  p = 0.15

     The mean is  \mu  =  5

Generally the distribution of sales call  made by a  telesales representative follows a binomial distribution  

i.e  

         X  \~ \ \ \  B(n , p)

and the probability distribution function for binomial  distribution is  

      P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

Generally the mean is mathematically represented as

     \mu =  n*  p

=>  5= n *  0.15

=>  n = 33

Generally the least number of calls that need to be made by a representative for the  probability of at least 1 sale to exceed 0.95 is mathematically represented as

      P( X \ge 1) = 1 - P( X < 1 ) > 0.95

=>    P( X \ge 1) = 1 - P( X =0 ) > 0.95

=>    P( X \ge 1) = 1 - [ ^{n}C_0 *  (0.15 )^0 *  (1- 0.15)^{n-0}] > 0.95

=>    1 - [1  *  1*  (0.85)^{n}] > 0.95

=>    [(0.85)^{n}] > 0.05

taking natural  log of both sides

n = \frac{ln(0.05)}{ln(0.85)}

=>  n = 19

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